3.5

Two technicians are discussing torque wrenches. Technician A says that a torque wrench is capable of tightening a fastener with

USPshnik [31]

Djfigud ticks fjyigud dhy

7 0

3 years ago

the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is

cluponka [151]

Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

Force ( F ) = 3.2 x 10^-9 N

Mass ( m1 ) = 40 kg

To find : Mass ( m2 ) = ?

Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

m2 = 0.8 x 10^-9 kg

3 0

3 years ago

Ahmed fills a basket with shopping weighing 10 kg. How much work is being done on the shopping basket when he lifts it verticall

NISA [10]

The work done by the shopping basket is 147 J.

Work is said to be done whenever a force moves an object through a certain distance.

The amount of work done on the shopping basket can be calculated using the formula below.

Formula:

W = mgh

Where:

W = Amount of work done by the basket
m = mass of the shopping basket
h = height of the shopping basket
g = acceleration due to gravity.

Form the question,

Given:

m = 10 kg
h = 1.5 m
g = 9.8 m/s²

Substitute these values into equation 2

W = 10(1.5)(9.8)
W = 147 J.

Hence, The work done by the shopping basket is 147 J.

Learn more about work done here: brainly.com/question/18762601

6 0

2 years ago

Sometimes referred to as the law of inertia. An object at rest stays at rest and an object in motionstays in motion with the sam

Lera25 [3.4K]

Answer:

D

Explanation:

Newtons first law states that if an object is at rest it will stay at rest only if an unbalanced force acts on it. As well as if an object is in motion it will stay in motion unless an unbalanced force acts on it.

Ps- The object will stay moving in the same speed and direction.

5 0

3 years ago

Given the isotope 2Fes, which has an actual mass of 55.934939 u: a) b) Determine the mass defect of the nucleus in atomic mass u

SSSSS [86.1K]

Answer:

Mass defect of each iron-56 nuclei:

The binding energy per nucleon of Iron-56 is approximately 8.6 MeV.

Explanation:

According to the physics constants table on Chemistry Libretexts:

Proton rest mass: $\rm 1.0072765\;amu$ ;
Neutron rest mass: $\rm 1.0086649\; amu$ .
Speed of light in vacuum: $\rm 2.99792458\times 10^{8}\;m\cdot s^{-1}$ .
Charge on an electron: $\rm 1.6021765\times 10^{-19}\;C$ .

The mass defect of a nucleus is equal to the sum of the mass of its parts (protons and, in most cases, neutrons) minus the mass of the nucleus.

The atomic number of iron is 26. There are 26 protons in each iron-56 nucleus. The mass number 56 indicates that there are 56 nucleons (neutrons and protons) in each iron-56 nucleus. The other $56 - 26 = 30$ particles are neutrons.

The mass of protons and neutrons in each iron-56 nucleus will be:

$\rm 26 \times 1.0072765 + 30 \times 1.0086649 = 56.464736\;amu$ .

According to this question, the mass of an iron-56 nucleus is equal to 55.934939 amu. The mass defect will be

$\rm 56.464736 - 55.934939 = 0.514197\;amu$ .

By the mass-energy equivalence,

$E = m\cdot c^{2}$ .

Refer to this equation, the speed of light in vacuum $c^{2}$ is the conversion factor between mass $m$ and energy $E$ . The value of $c$ is usually given only in SI units $\rm m\cdot s^{-1}$ . Accordingly, the value of $c^{2}$ will be in the SI unit $\rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}$ .

Convert million electron-volts to joules.

One electron-volt is equal to the electrical work done moving an electron across a potential difference of one volt.

$\begin{aligned}\rm 1 MeV&= \rm 10^{6}\; eV\\ &= \rm (10^{6}\times 1.6021765\times 10^{-19}\;C)\times 1\; V\\&=\rm 1.6021765\times 10^{-19}\;J\end{aligned}$ .

Convert the unit of $c^{2}$ from $\rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}$ to the desired $\rm MeV \cdot amu^{-1}$ :

$\begin{aligned}c^{2} &= \rm {\left(2.99792458\times 10^{8}\;m\cdot s^{-1}\right)}^{2}\\&=\rm 8.987551787\times 10^{16}\; m^{2}\cdot s^{-2}\\ &= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\\&= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\times \frac{1\;MeV}{1.6021765\times 10^{-13}\;J}\times \frac{1\times 10^{-3}\;kg}{6.022142\times 10^{23}\;amu}\\&\approx \rm 931.602164\;MeV\cdot amu^{-1}\end{aligned}$ .

Total binding energy in each iron-56 nucleus:

$\begin{aligned}E &= m\cdot c^{2}\\&= \rm 0.514197\;amu \times 9.31602164\;MeV\cdot amu^{-1} \\&=\rm 479.027038\; MeV \end{aligned}$ .

Again, the mass number 56 indicates that there are 56 nucleons in each iron-56 nucleus. The binding energy per nucleon of iron-56 $\mathrm{^{56}Fe}$ will be:

$\displaystyle \rm \frac{479.027038\; MeV}{56} \approx 8.6\; MeV$ .

6 0

3 years ago